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The exponentiated Hencky energy: anisotropic extension and case studies

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Abstract

In this paper we propose an anisotropic extension of the isotropic exponentiated Hencky energy, based on logarithmic strain invariants. Unlike other elastic formulations, the isotropic exponentiated Hencky elastic energy has been derived solely on differential geometric grounds, involving the geodesic distance of the deformation gradient \({{\varvec{F}}}\) to the group of rotations. We formally extend this approach towards anisotropy by defining additional anisotropic logarithmic strain invariants with the help of suitable structural tensors and consider our findings for selected case studies.

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Notes

  1. Note that not every objective and isotropic energy function can be expressed in terms of the logarithmic strain measures alone, see Neff et al. [48], whereas every such energy can be expressed in terms of the logarithmic strain tensor \(\log \varvec{U}\).

  2. In the contracted notation the tensorial indices are allocated to the matrix indexes as follows \(\{11,22,33,12,23,13\}\rightarrow \{1,2,3,4,5,6\}\).

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Acknowledgements

The first two authors gratefully acknowledge support by the Deutsche Forschungsgemeinschaft in the Priority Program 1748 under the project “Novel finite elements for anisotropic media at finite strain” (SCHR 570/23-1), Sects. 14.3. Further, the first two authors would like to acknowledge support by the Deutsche Forschungsgemeinschaft within the framework of the project “Domain-decomposition-based fluid structure interaction algorithms for highly nonlinear and anisotropic elastic arterial wall models in 3D“ (SCHR 570/15-2) under the D-A-CH agreement, Sects. 4.45.

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Appendix

Appendix

1.1 Notes on the Hencky tensor

The Hencky strain tensor is defined through

$$\begin{aligned} \log {{\varvec{U}}}= \frac{1}{2} \log {{\varvec{C}}}\end{aligned}$$
(74)

and

$$\begin{aligned} \mathrm{tr} (\log {{\varvec{U}}}) = \log (\mathrm{det} {{\varvec{U}}}). \end{aligned}$$
(75)

The symmetric right Cauchy–Green tensor \({{\varvec{C}}}\) in spectral decomposition is given by

$$\begin{aligned} {{\varvec{C}}}= \sum _{k=1}^3 \widehat{\lambda }_k {{\varvec{N}}}^k\otimes {{\varvec{N}}}^k = \sum _{k=1}^3 \widehat{\lambda }_k {{\varvec{P}}}^k,\quad \text{ with }\quad {{\varvec{P}}}^k = {{\varvec{N}}}^k\otimes {{\varvec{N}}}^k \end{aligned}$$
(76)

and for the Hencky strain we obtain

$$\begin{aligned} \log {{\varvec{U}}}= & {} \sum _{k=1}^3 \frac{1}{2}\log (\widehat{\lambda }_k) {{\varvec{N}}}^k\otimes {{\varvec{N}}}^k \nonumber \\= & {} \sum _{k=1}^3 \frac{1}{2}\log (\widehat{\lambda }_k) {{\varvec{P}}}_k,\quad \hbox {with}\quad {{\varvec{P}}}_k = {{\varvec{N}}}^k\otimes {{\varvec{N}}}^k . \end{aligned}$$
(77)

The first derivative of \(\log {{\varvec{U}}}\) with respect to \({{\varvec{C}}}\) can be computed as

$$\begin{aligned} \frac{\partial {\log {{\varvec{U}}}}}{\partial {{{\varvec{C}}}}} =&\sum _{k=1}^3 {{\varvec{P}}}_k\otimes \frac{\partial {\frac{1}{2}\log \widehat{\lambda }_k}}{\partial {{{\varvec{C}}}}} + \frac{1}{2}\log \widehat{\lambda }_k \frac{\partial {{{\varvec{P}}}_k}}{\partial {{{\varvec{C}}}}} \nonumber \\ =&\sum _{k=1}^3 {{\varvec{P}}}_k\otimes \frac{\partial {\frac{1}{2}\log \widehat{\lambda }_k}}{\partial {\widehat{\lambda }_k}} \, \frac{\partial {\widehat{\lambda }_k}}{\partial {{{\varvec{C}}}}} + \frac{1}{2}\log \widehat{\lambda }_k \frac{\partial {{{\varvec{P}}}_k}}{\partial {{{\varvec{C}}}}} \nonumber \\ =&\sum _{k=1}^3 \frac{1}{2}\widehat{\lambda }_k^{-1}{{\varvec{P}}}_k\otimes \frac{\partial {\widehat{\lambda }_k}}{\partial {{{\varvec{C}}}}} + \frac{1}{2}\log \widehat{\lambda }_k \frac{\partial {{{\varvec{P}}}_k}}{\partial {{{\varvec{C}}}}} \end{aligned}$$
(78)

Considering that

$$\begin{aligned} \frac{\partial {\widehat{\lambda }_k}}{\partial {{{\varvec{C}}}}}= & {} {{\varvec{P}}}_k\qquad \hbox {and} \qquad \nonumber \\ \frac{\partial {{\varvec{P}}}_k }{\partial {{\varvec{C}}}}= & {} \sum _{j=1, j\ne k}^3 \frac{{{\varvec{P}}}_k\boxtimes {{\varvec{P}}}_j^T + {{\varvec{P}}}_j\boxtimes {{\varvec{P}}}_k^T}{\widehat{\lambda }_{k}-\widehat{\lambda }_{j}}, \end{aligned}$$
(79)

see for instance Jog [33], we find that

$$\begin{aligned} \frac{\partial \log {{\varvec{U}}}}{\partial {{\varvec{C}}}}= & {} \sum _{k=1}^3 \frac{1}{2}\widehat{\lambda }_{k}^{-1} {{\varvec{P}}}_k \otimes {{\varvec{P}}}_k \nonumber \\&+ \sum _{k=1}^3 \sum ^3_{\begin{array}{c} j=1\\ k\ne j \end{array}} \frac{\frac{1}{2}(\log \widehat{\lambda }_k) -\frac{1}{2}(\log \widehat{\lambda }_j)}{\widehat{\lambda }_{k}-\widehat{\lambda }_{j}}\nonumber \\&({{\varvec{P}}}_k \boxtimes {{\varvec{P}}}_j + {{\varvec{P}}}_j\boxtimes {{\varvec{P}}}_k). \end{aligned}$$
(80)

The second derivative for the linearization is given by

$$\begin{aligned} \frac{\partial ^2 \log {{\varvec{U}}}}{\partial {{\varvec{C}}}\partial {{\varvec{C}}}} =&\sum _{k=1}^3 {{\varvec{P}}}_k\otimes {{\varvec{P}}}_k\otimes \frac{\partial {\frac{1}{2}\widehat{\lambda }_{k}^{-1}}}{\partial {{{\varvec{C}}}}} \nonumber \\&+ \frac{1}{2}\widehat{\lambda }_{k}^{-1}\left[ \left( \frac{\partial {{\varvec{P}}}_k}{\partial {{\varvec{C}}}}\otimes {{\varvec{P}}}_k\right) ^{\begin{array}{c} 35\\ T \end{array}\begin{array}{c} 46\\ T \end{array}} + {{\varvec{P}}}_k\otimes \frac{\partial {{\varvec{P}}}_k}{\partial {{\varvec{C}}}}\right] \nonumber \\&+ \frac{1}{2}(\mathrm{log}\widehat{\lambda }_k) \frac{\partial ^2 {{\varvec{P}}}_k}{\partial {{\varvec{C}}}\partial {{\varvec{C}}}} + \frac{\partial {{{\varvec{P}}}_k}}{\partial {{{\varvec{C}}}}} \otimes \frac{\partial {\frac{1}{2}\widehat{\lambda }_k^{-1}}}{\partial {{{\varvec{C}}}}} \nonumber \\ =&\sum _{k=1}^3 -\frac{1}{2}\widehat{\lambda }_{k}^{-2} {{\varvec{P}}}_k\otimes {{\varvec{P}}}_k\otimes {{\varvec{P}}}_k \nonumber \\&+ \frac{1}{2}\widehat{\lambda }_k^{-1}\left[ \left( \frac{\partial {{\varvec{P}}}_k}{\partial {{\varvec{C}}}}\otimes {{\varvec{P}}}_k\right) ^{\begin{array}{c} 35\\ T \end{array}\begin{array}{c} 46\\ T \end{array}} + {{\varvec{P}}}_k\otimes \frac{\partial {{\varvec{P}}}_k}{\partial {{\varvec{C}}}}\right] \nonumber \\&+ \frac{1}{2}(\mathrm{log}\widehat{\lambda }_k) \frac{\partial ^2 {{\varvec{P}}}_k}{\partial {{\varvec{C}}}\partial {{\varvec{C}}}} + \frac{1}{2} \widehat{\lambda }_k^{-1} \frac{\partial {{{\varvec{P}}}_k}}{\partial {{{\varvec{C}}}}} \otimes {{\varvec{P}}}_k, \end{aligned}$$
(81)

where

$$\begin{aligned} \frac{\partial ^2 {{\varvec{P}}}_k}{\partial {{\varvec{C}}}\partial {{\varvec{C}}}} =&\sum ^3_{\begin{array}{c} j=1\\ k\ne j \end{array}} \frac{1}{\widehat{\lambda }_{k}-\widehat{\lambda }_{j}}\nonumber \\&\left[ \left( {{\varvec{P}}}_j^T \otimes \left( \frac{\partial {{{\varvec{P}}}_k}}{\partial {{{\varvec{C}}}}} \right) ^{\begin{array}{c} 12\\ T \end{array}}\right) ^{\begin{array}{c} 23\\ T \end{array}} + \left( {{\varvec{P}}}_k\otimes \frac{\partial {{{\varvec{P}}}_j}}{\partial {{{\varvec{C}}}}} \right) ^{\begin{array}{c} 23\\ T \end{array}} \right. \nonumber \\&\left. + \left( {{\varvec{P}}}_k^T \otimes \left( \frac{\partial {{{\varvec{P}}}_j}}{\partial {{{\varvec{C}}}}} \right) ^{\begin{array}{c} 12\\ T \end{array}}\right) ^{\begin{array}{c} 23\\ T \end{array}} + \left( {{\varvec{P}}}_j\otimes \frac{\partial {{{\varvec{P}}}_k}}{\partial {{{\varvec{C}}}}} \right) ^{\begin{array}{c} 23\\ T \end{array}} \right] \nonumber \\&+ \frac{1}{\left( \widehat{\lambda }_k -\widehat{\lambda }_j\right) ^2} \left( {{\varvec{P}}}_k\boxtimes {{\varvec{P}}}_j + {{\varvec{P}}}_j \boxtimes {{\varvec{P}}}_k\right) \nonumber \\ {}&\otimes \left( {{\varvec{P}}}_j-{{\varvec{P}}}_k\right) . \end{aligned}$$
(82)

The exponent and the logarithm of an arbitrary symmetric tensor may also be be expressed with help of a Taylor expansion of the form

$$\begin{aligned} \mathrm{exp}(\bullet )&= \mathbf{1}+ \sum _{m =1}^\infty \frac{1}{m!}(\bullet )^m\qquad \hbox {and} \end{aligned}$$
(83)
$$\begin{aligned} \mathrm{log}(\bullet )&=\sum _{k=1}^\infty \frac{-1^{k-1}}{k}[(\bullet )-\mathbf{1}]^k, \end{aligned}$$
(84)

where the latter is convergent in a neighborhood of \(\mathbf{1}\).

1.2 Conjugate stress tensors

The following considerations are adapted from Ogden [50].

The constitutive equation for the stresses are derived form the (isothermal) entropy inequality

$$\begin{aligned} \dot{\psi } - \langle {{\varvec{P}}},\dot{{{\varvec{F}}}}\rangle \ge 0, \end{aligned}$$
(85)

From the latter we deduce the constitutive relation \({{\varvec{P}}}= \partial _{{{\varvec{F}}}}\psi \). Let the generalized Lagrangean strain measures

$$\begin{aligned} {{\varvec{E}}}^{(m)} = {\left\{ \begin{array}{ll} \frac{1}{2}\left( {{\varvec{U}}}^{m}-\mathbf{1}\right) \quad &{} m\ne 0 \\ \frac{1}{2}\mathrm{log}\,{{\varvec{U}}}\quad &{}m=0 \end{array}\right. } \end{aligned}$$
(86)

and Eulerian strain measures

$$\begin{aligned} {{\varvec{K}}}^{(m)} = {\left\{ \begin{array}{ll} \frac{1}{m}\left( {{\varvec{V}}}^{m}-\mathbf{1}\right) \quad &{} m\ne 0 \\ \mathrm{log}{{\varvec{V}}}\quad &{}m=0 \end{array}\right. } \end{aligned}$$
(87)

be given, we aim to find the corresponding constitutive equations. The so called stress power may be written as

$$\begin{aligned} \langle {{\varvec{P}}},\dot{{{\varvec{F}}}}\rangle= & {} \langle {\varvec{\tau }},{{\varvec{D}}}\rangle \nonumber \\= & {} \langle {{\varvec{S}}},\dot{{{\varvec{E}}}}\rangle \nonumber \\= & {} \langle \partial _{{{\varvec{E}}}^{(m)}}\psi ,\dot{{{\varvec{E}}}}^{(m)}\rangle =\langle \partial _{{{\varvec{K}}}^{(m)}}\psi ,\dot{{{\varvec{K}}}}^{(m)}\rangle , \end{aligned}$$
(88)

where \({{\varvec{D}}}= \frac{1}{2}({{\varvec{L}}}+{{\varvec{L}}}^T)\) and \({{\varvec{L}}}= {\text {grad}}\dot{{{\varvec{x}}}}\). Considering that \(\dot{{{\varvec{E}}}}=\frac{1}{2}\left( \dot{{{\varvec{F}}}}^T{{\varvec{F}}}+{{\varvec{F}}}^T\dot{{{\varvec{F}}}}\right) \), we obtain the relations

$$\begin{aligned} {{\varvec{P}}}= {{\varvec{F}}}{{\varvec{S}}}= {\varvec{\tau }}{{\varvec{F}}}^{-T}. \end{aligned}$$
(89)

The pairs in Eq. (88) are said to be work conjugate. By making use of the fact that \({{\varvec{R}}}^T\dot{{{\varvec{R}}}}= -\dot{{{\varvec{R}}}}^T{{\varvec{R}}}\), we may rewrite \(\dot{{{\varvec{E}}}}= \frac{1}{2}\left( {{\varvec{U}}}\dot{{{\varvec{U}}}}+\dot{{{\varvec{U}}}}{{\varvec{U}}}\right) \) and we are able to reformulate

$$\begin{aligned} \Big \langle {{\varvec{S}}},\frac{1}{2}\left( {{\varvec{U}}}\dot{{{\varvec{U}}}}+\dot{{{\varvec{U}}}}{{\varvec{U}}}\right) \Big \rangle = \Big \langle \underbrace{\frac{1}{2}\left( {{\varvec{S}}}{{\varvec{U}}}+{{\varvec{U}}}{{\varvec{S}}}\right) }_{{{\varvec{T}}}_\mathrm{Biot}},\dot{{{\varvec{U}}}}\Big \rangle , \end{aligned}$$
(90)

such that we directly obtain the Biot stress \({{\varvec{T}}}_\mathrm{Biot}=\partial _{{{\varvec{U}}}}\psi ^\#({{\varvec{U}}})\), work conjugate to \({{\varvec{U}}}\) from the entropy inequality. With \({{\varvec{U}}}= {{\varvec{R}}}^T{{\varvec{V}}}{{\varvec{R}}}\) we are able to relate \({{\varvec{E}}}^{(m)}\) and \({{\varvec{K}}}^{(m)}\) and the corresponding time derivatives as follows:

$$\begin{aligned} {{\varvec{E}}}^{(m)}= & {} {{\varvec{R}}}^T{{\varvec{K}}}^{(m)}{{\varvec{R}}}\quad \hbox {and} \quad \nonumber \\ \dot{{{\varvec{E}}}}^{(m)}= & {} {{\varvec{R}}}^T\dot{{{\varvec{K}}}}^{(m)}{{\varvec{R}}}+ {{\varvec{E}}}^{(m)}{{\varvec{R}}}^T\dot{{{\varvec{R}}}} -{{\varvec{R}}}^T\dot{{{\varvec{R}}}}{{\varvec{E}}}^{(m)}. \end{aligned}$$
(91)

Regarding the generalized stress-power it follows

$$\begin{aligned}&\langle \partial _{{{\varvec{E}}}^{(m)}}\psi ,\dot{{{\varvec{E}}}}^{(m)}\rangle \nonumber \\&\quad = \langle {{\varvec{R}}}(\partial _{{{\varvec{E}}}^{(m)}}\psi ){{\varvec{R}}}^T,\dot{{{\varvec{K}}}}^{(m)}\rangle \nonumber \\&\qquad +\left\langle \left[ (\partial _{{{\varvec{E}}}^{(m)}}\psi ){{\varvec{E}}}^{(m)}-{{\varvec{E}}}^{(m)} (\partial _{{{\varvec{E}}}^{(m)}}\psi )\right] ,{{\varvec{R}}}^T\dot{{{\varvec{R}}}}\right\rangle \end{aligned}$$
(92)

Only if \({{\varvec{E}}}^{(m)}\partial _{{{\varvec{E}}}^{(m)}}\psi = \partial _{{{\varvec{E}}}^{(m)}}\psi {{\varvec{E}}}^{(m)}\), i.e. \(\partial _{{{\varvec{E}}}^{(m)}}\psi \) is coaxial with \({{\varvec{E}}}^{(m)}\), it immediately follows that the constitutive law results in

$$\begin{aligned} \frac{\partial {\psi }}{\partial {{{\varvec{K}}}^{(m)}}} = {{\varvec{R}}}(\partial _{{{\varvec{E}}}^{(m)}}\psi ){{\varvec{R}}}^T \end{aligned}$$
(93)

and the stress power is expressed through \(\langle \dot{{{\varvec{K}}}}^{(m)}, {{\varvec{R}}}(\partial _{{{\varvec{E}}}^{(m)}}\psi ){{\varvec{R}}}^T\rangle \), which is identical to Eq. (88).

The case that \(\partial _{{{\varvec{E}}}^{(m)}}\psi \) is coaxial with \({{\varvec{E}}}^{(m)}\) implies that also \(\partial _{{{\varvec{E}}}^{(m)}}\psi \) and \({{\varvec{U}}}\) are coaxial. Under this assumption one may show that

(94)

and it follows that

(95)

for isotropic materials. Inserting the latter result in Eq. (93) we obtain the relation

$$\begin{aligned} {\varvec{\tau }}= \frac{\partial {\psi (\mathrm{log}{{\varvec{V}}})}}{\partial {\mathrm{log}{{\varvec{V}}}}} . \end{aligned}$$
(96)

Regarding the conjugate stress to \(\log {{\varvec{U}}}\) the reader is also referred to Hill [26, 27], Hoger [28].

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Schröder, J., von Hoegen, M. & Neff, P. The exponentiated Hencky energy: anisotropic extension and case studies. Comput Mech 61, 657–685 (2018). https://doi.org/10.1007/s00466-017-1466-4

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